Lecture 1. Integration of fractional-rational functions (part 1)
Integration of fractional-rational functions.
This lecture is devoted to an extremely important topic - integration of fractional-rational functions. First, let's formulate a definition. The fractional-rational function is a function that is represented as a ratio of two polynomials. Note, the symbols Pn(x) and Qm(x) designate the polynomials of the variable x, the indices denote the degree of the polynomial. There are proper and improper fractions, we compare the degrees of the polynomials: if the degree of numerator less than degree of denominator, then the fraction is proper, if it is more or equal, then the fraction is improper. The most common examples.
The functions f and g are samples of improper fractions. Note that for the function f, the degree of the numerator is four, the denominator is three, and four is greater than three, so, the fraction is improper. The function g has a third - degree polynomial in the numerator as well as in the denominator, the degrees are equal; the fraction is also called improper. And now the example of the proper fraction. The function h is proper because the degree of the numerator is the first degree, and it is less than the degree of the denominator which is the second degree.
Now we formulate an important statement that we are going to use when integrating fractional-rational functions.
Any improper fractional-rational function can be represented as the sum of a polynomial and a regular fraction.
In order to implement this representation, we use two main methods: they are column division with remainder and identical transformations of this function. Let's look at each of these methods using an example. So, the function f(x), we have considered it, is an improper fraction. We divide by a column in the same way as we do it for numbers. We look at the highest powers: 2x^4 for the function is in the numerator and x^3 is in the denominator.
When dividing, we get 2x, write it under the line, we multiply the polynomial x^3-x+2 by 2x and write this in the same way as we do it with numbers. At the next stage, we perform subtraction. Please note, students often make mistakes in this particular place. Note that 2x^2 remains with the plus sign, since here there is minus -2x^2, the sign changes. Well, as a result, we understand that 2x^2 must be divided by x^3, the power is less than the third, the division stops there. So, 2x is the integer part, and the remainder is what we got under the line. As a result, the function f(x) is represented as the sum of the polynomial and the proper fraction. The numerator contains the remainder, and the denominator is similar to the one in the function f.
We perform identical transformations for the second function g(x), considered before, which is improper. We represent the numerator in such a way that the denominator x^3+2 appears in parentheses. After that, we have to perform the division of each summand of the numerator, which we call the resulting divisions. As a result, we get a representation of the function g as the sum of a polynomial and a proper fraction.
So let's analyze the situation. If we have an improper fraction, then it is represented as the sum of the polynomial and the proper fraction. Next step. How do we integrate the proper fraction? It turns out that any proper fractional-rational function can be represented as a sum of the simple fractional-rational functions.
Briefly, we will talk about rational fractions. The situation at this stage looks like this. An improper fraction is represented as the sum of a polynomial and a proper fraction, which is in turn the sum of the simplest fractions. A polynomial is an integrable function. We have to understand what the simplest fractions are and how they are integrated.
The simple fractional-rational functions are functions of the following type. There are only four types of elementary fractions. The first and second types differ only in the power of the denominator. A (an upper-case), a (a lower-case), and the symbols p and q are found here - these are real numbers. The first and second types are the number A divided by x-a having some power. The first type is characterized by the first power, if the power is higher, it is the second type.
Similarly, the third and fourth types differ only in the power of the denominator - the denominator is a square trinomial, the discriminant is negative. For the fourth type, the square trinomial is greater than one, and for the third type, it is the first power.
The next step we need to take is to understand how to integrate these fractions.
Let’s consider the elementary fraction of the first type. We are going to calculate the integral of this function. The first step is to place the number A out of the integral sign, we do it using the properties of the indefinite integral. The next step is to enter x - a under the differential sign; as a result we get a table integral, where there is an expression x-a instead of the variable x in the table of integrals. We get the answer.
Let’s consider the elementary fraction of the second type. Integration is carried out in exactly the same way. The only thing that differs it from the first type is that we apply the formula of the integral from a power function in the integrals table. The next step is to write down the answer.
Let’s consider the third type of elementary fractions. In order to make it more clear to you, we will analyze the integration of such a fraction using an example. This is the elementary fraction of the third type. Note that the numerator has a linear expression ax+b (2 and -3 here are the numbers a and b), and the denominator has a square trinomial with a negative discriminant. This is the elementary fraction of the third type. For integration, we are going to consider the denominator (this is a square trinomial) and select the full square. This conversion is a part of the school curriculum, so I think you can handle it easily.
The next step is to assign the new variable t to the expression that appears in parentheses and is squared.
As a result, this variable t and the square trinomial are converted to the expression t^2+1. So, the situation with the denominator of the original fraction is clear. We need to understand what 2x-3 in the integral and the expression dx (the differential of the variable) are equal to. In order to do it, we will find what x equals to if we substitute the variable. x is t-2. After that, the numerator (we substitute t-2 instead of x) is converted to 2t-7, and the differential is calculated using the formula. What is a differential? This is the derivative of the function t-2 multiplied by the differential of the variable t. The derivative is equal to one. As a result, we get dx is equal to dt. We need to replace the integral for a new variable t, and the next step is to perform a slow division of the numerator by the denominator.
We convert the difference integral to the difference of integrals of integrals and take out a constant multiplier for the sign of the integral, using the linearity property of the indefinite integral. Note that the second integral is a tabular one, we won’t use it for a while, and for the first integral we will perform a transformation, which is called introducing a function under the sign of the differential. Note that tdt is the differential of the denominator with a coefficient of 1/2. So what do we have? In the first integral, we have to place the coefficient 1/2 out of the sign of the integral, and as a result, both integrals are tabular. We calculate them using the table of integrals, and we have to replace the variable t - do not forget to return to the original variable – a reverse replacement, remember where we started. And it is not necessary to substitute t^2 for (x+2)^2 instead. Note that t^2+1 is the square trinomial. We started with it. And we get the final answer.
Now we have to consider the elementary fraction of the fourth type. Just as the second type differs from the first one, so the fourth type differs from the third one. Therefore, the same transformations are carried out here: selecting a full square, and replacing a variable. The only difficulty is that we face an integral of the form and with the index n, note how it is presented; we use a recurrent formula to calculate it.
We will not prove this formula here.
We will study how you can use this integral formula. Let’s take an example. Let’s use the integral In, which is set in the recurrent formula. Note that this is indeed an integral where a^2 is one and n is three.
Let's use this formula. So, I3 (I with index 3) is a given integral where a^2 is one, n is equal to three. We insert these values into the recurrent formula, perform the conversion, and get this result.
The next step is to calculate the integral I2 (I with index 2). All values are the same here except for n, which is equal to two, this is the exponent. We use the recurrent formula and get the integral I1.
The integral I1 is the same integral with the index In (I with index n), only n is equal to one. This is already a tabular integral; we find it in the table of integrals, so it can be calculated. We get the answer for the integral I2. Let's look at the top entry of I3. The integral I3 is expressed in terms of the integral I2, we have obtained the value for it. We return to the integral I3, substituting the obtained value for I2. This further transformation leads us to the answer.
Write down the answer.